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What is the partial derivative of $$\frac{\partial x}{\partial y}$$

when $x$ and $y$ are a part of a function $f(x,y)$?

Using an example of:

$$f(x,y) = x+y$$ Given the definition of holding all else constant and varying x with respect to y (thus setting the function to zero), would it be corrrect to say that the partial derivative is simply negative one?

One motivation for this is the formula (note that this is just re-arranged formula for total derivative or differential):

$$\frac{dy}{dx} = -\frac {\partial F /\partial x}{\partial F / \partial y} +\frac{dF }{dx }\frac{\partial y}{\partial F}$$

And whether the Fs in the 2nd term can in fact be factored out. If that is the case, then:

$$\frac{\partial x}{\partial y} = \frac {\partial F /\partial x}{\partial F / \partial y} =\frac{dF }{dx }\frac{\partial y}{\partial F}-\frac{dy}{dx} $$

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  • $\begingroup$ is here $y=y(x)$? $\endgroup$ Jun 26 '16 at 17:25
  • $\begingroup$ Proper notation is $\dfrac{\partial x}{\partial y},$ not $\dfrac{\delta x}{\delta y}.$ $\qquad$ $\endgroup$ Jun 26 '16 at 17:27
  • $\begingroup$ Note that for $f(x,y)=x+y$, we have $$\begin{align} df&=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy\\\\ &=dx+dy \end{align}$$ If $y=g(x)$, then $dy=g'(x)\,dx$ and we find that $$\frac{df}{dx}=1+g'(x)$$ $\endgroup$
    – Mark Viola
    Jun 26 '16 at 17:28
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    $\begingroup$ When you say "the partial derivative", you have to clarify WRT what. $\endgroup$
    – GFauxPas
    Jun 26 '16 at 18:04
  • $\begingroup$ Taking up GFauxPas, if you have some function $f(x,y)$, when you write $\frac{\partial f}{\partial x}$ you should really write it as $(\frac{\partial f}{\partial x})_{y=\text{constant}}$. $\endgroup$
    – jim
    Jun 26 '16 at 18:54
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Partial derivatives are defined in terms of functions. Total derivatives are defined in terms of variables. They are the same concept defined in 2 different languages.

Partial derivatives, when used correctly, are always the derivative of a function with respect to one of it's parameters. So, for example, suppose you are given to assume:

$$f(x, y) = x^2y + \sin(x)$$

And asked to calculate $\frac{\partial f}{\partial x}$. Here, you see that the denominator is $x$, and so you look for that in the definition of $f$, see that it is the first parameter, and take the derivative of $f$ with respect to that first parameter. So

$$\frac{\partial f}{\partial x} = 2xy + \cos(x)$$

$\frac{\partial f}{\partial z}$ in this case is meaningless, because $z$ is not used as one of the parameters defining $f$. Suppose you were asked to calculate

$$\frac{\partial f(z, {\color{red} x}^2)}{\partial {\color {green} x}}$$

First, realize that the red $x$ and the green $x$ represent 2 different things. The red $x$ is a variable, the green $x$ represents the parameter used to define $f$. They are not interchangable, it is rude to use the same name for 2 different concepts, but it is not too uncommon.

Second, the above is a shorthand for $\left(\frac{\partial f}{\partial x}\right)(z, x^2)$, that is, take the derivative then apply the arguments. It's value is $2zx^2 + \cos(z)$.

What is the partial derivative of $$\frac{\partial x}{\partial y}$$

when $x$ and $y$ are a part of a function $f(x,y)$?

It is never correct to take the partial derivative of a variable with respect to another variable, or to take the partial derivative of a parameter with respect to another parameter. It is only meaningly to take the partial derivative of a function with respect to one of it's parameters. It's should be mentioned that referring to a parameter outside of the definition of a function is an abuse of notation, the name is not in scope at that point.

A total derivative is always of a variable with respect to another variable. Sometimes, a total derivative will be written as the derivative of a function with respect to a variable, but there the function is just a shorthand for "the variable representing the output of a function".

Suppose you are given

$$f(x) = x^2$$

And asked to calculate $\frac{{\rm d}f}{{\rm d}x}$. That is not defined. The $x$ in the definition is a parameter, the $x$ in the denominator is a variable.

On the other hand, what if you are asked to calculate $\frac{{\rm d}f(x)}{{\rm d}x}$, then both the previous $x$s represent variables. To evaluate this, you apply the argument, then calculate the derivative:

$$\frac{{\rm d}f(x)}{{\rm d}x}$$ $$\frac{{\rm d}x^2}{{\rm d}x}$$ $$2x$$

Note that the is the opposite order of what is done with partial derivatives.

With a partial derivative, you first calculate the derivative, then apply the arguments. With a total derivative, you first apply the arguments, then calculate the derivative.

Total derivatives are the type of notation you would expect a scientist to use, because the variables represent concepts like time, temperature, gravitational force, etc. Partial derivatives are what purists use, because functions are very easy to define concepts and make it easier to connect differential calculus to formal foundations of mathematics.

It is common to see these notations confused and used incorrectly even in college textbooks.

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  • $\begingroup$ It seems to me that many things in your answer are not really common among mathematicians. To me, $\frac{d f(x^2)}{dx}$ simply means "compute $df$ and then evaluate it at $x^2$. $\endgroup$
    – Siminore
    Jun 26 '16 at 19:18
  • $\begingroup$ Suppose $f(x) = x$. What do you think $\frac{{\rm d}f(x^2)}{{\rm d}x}$ should be? Or to put it another way, if $y = x^2$, what should $\frac{{\rm d}y}{{\rm d}x}$ be? $\endgroup$
    – DanielV
    Jun 26 '16 at 19:23
  • $\begingroup$ One, I suppose. $\endgroup$
    – Siminore
    Jun 26 '16 at 19:26
  • $\begingroup$ @Siminore It's only 1 with partial derivatives. You are implicitly associating the $x$ in the denominator with the parameter. Suppose you are given $g(z) = z$. What is $\frac{{\rm d}g(x^2)}{{\rm d}x}$ ? $\endgroup$
    – DanielV
    Jun 26 '16 at 19:29
  • $\begingroup$ The way I interpret $\frac{df (x^2)}{dx}$ is that $x^2$ is an inner function in the sense of function concatenation and chain rule, like we could write $\frac{df(g(x))}{dx}, g(x) = x^2$. To be pedantic one should require a set of parentheses after the d in the numerator. Like this $\frac{d\{f (x^2)\}}{dx}\left(x_0\right)$ to be clear what expression the differential operator should take on and what is the function value it should be evaluated at $(x_0)$. This way we can express Siminores interpretation like this $\frac{d\{f\}}{dx}(x^2)$. $\endgroup$ Jun 26 '16 at 20:18
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If in an $n$-dimensional "environment" $\Omega$ an $n$-tuple $(x_1,x_2,\dots, x_n)$ of coordinate functions has been selected then functions $f:\>\Omega\to{\mathbb R}$ appear as "functions of $n$ variables", and one denotes the value of $f$ at ${\bf p}=(p_1,\ldots,p_n)$ by $f(p_1,\ldots,p_n)$. It is only in such a situation that it makes sense to talk about partial derivatives ${\partial f\over\partial x_k}$ which then themselves are functions $\Omega\to{\mathbb R}$.

In the situation at hand we have the agreed on coordinate functions $x:\>{\mathbb R}^2\to{\mathbb R}$, and similarly $y$. Therefore any real-valued function $f$ defined on some open domain $\Omega\subset{\mathbb R}^2$ may have partial derivatives ${\partial f\over\partial x}$ and ${\partial f\over\partial y}$. In particular this is the case for the coordinate functions $x$ and $y$ themselves. Thinking about it one comes to the conclusion that $${\partial x\over\partial x}\equiv1,\qquad{\partial x\over \partial y}\equiv0\ ,$$ and similarly for $y$.

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